二十六角形

正二十六角形においては、中心角と外角は13.846…°で、内角は166.153…°となる。一辺の長さが a の正二十六角形の面積 S は

${\displaystyle S={\frac {26}{4}}a^{2}\cot {\frac {\pi }{26}}\simeq 53.53232a^{2}}$

${\displaystyle \cos(2\pi /26)}$を平方根と立方根で表すと

${\displaystyle \cos {\frac {2\pi }{26}}=\cos {\frac {\pi }{13}}={\frac {1}{12}}{\sqrt {72+72\cdot \cos {\frac {2\pi }{13}}}}={\frac {1}{12}}{\sqrt {72+72\cdot {\frac {1}{12}}\left({\sqrt[{3}]{104-20{\sqrt {13}}+12{\sqrt {-39}}}}+{\sqrt[{3}]{104-20{\sqrt {13}}-12{\sqrt {-39}}}}+{\sqrt {13}}-1\right)}}=0.970941...}$

関係式

${\displaystyle {\begin{aligned}&\alpha =2\cos {\frac {2\pi }{26}}+2\cos {\frac {6\pi }{26}}+2\cos {\frac {18\pi }{26}}={\frac {1+{\sqrt {13}}}{2}}\\&\beta =2\cos {\frac {14\pi }{26}}+2\cos {\frac {10\pi }{26}}+2\cos {\frac {22\pi }{26}}={\frac {1-{\sqrt {13}}}{2}}\\\end{aligned}}}$

三次方程式の係数を求めると

${\displaystyle {\begin{aligned}&2\cos {\frac {2\pi }{26}}\cdot 2\cos {\frac {6\pi }{26}}+2\cos {\frac {6\pi }{26}}\cdot 2\cos {\frac {18\pi }{26}}+2\cos {\frac {18\pi }{26}}\cdot 2\cos {\frac {2\pi }{26}}=-1\\&2\cos {\frac {2\pi }{26}}\cdot 2\cos {\frac {6\pi }{26}}\cdot 2\cos {\frac {18\pi }{26}}=\beta -2\end{aligned}}}$

解と係数の関係より

${\displaystyle x^{3}-\alpha x^{2}-x-(\beta -2)=0}$

変数変換

${\displaystyle x=y+\alpha /3}$

整理すると

${\displaystyle y^{3}-{\frac {13+{\sqrt {13}}}{6}}y+{\frac {26+5{\sqrt {13}}}{27}}=0}$

三角関数、逆三角関数を用いて解は

${\displaystyle x={\frac {1+{\sqrt {13}}}{6}}+{\frac {2}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}\cos \left({\frac {1}{3}}\arccos {\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}\right)}$

平方根、立方根を用いて

${\displaystyle x={\frac {1+{\sqrt {13}}}{6}}+{\frac {1}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}+i{\frac {3{\sqrt {39}}}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}}}+{\frac {1}{3}}{\sqrt {\frac {13+{\sqrt {13}}}{2}}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}-i{\frac {3{\sqrt {39}}}{2\left({\frac {13+{\sqrt {13}}}{2}}\right)^{\tfrac {3}{2}}}}}}}$

${\displaystyle x={\frac {1+{\sqrt {13}}}{6}}+{\frac {1}{3}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}+i{\frac {3{\sqrt {39}}}{2}}}}+{\frac {1}{3}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}-i{\frac {3{\sqrt {39}}}{2}}}}}$

${\displaystyle \cos(2\pi /26)}$を平方根と立方根で表すと

${\displaystyle \cos {\frac {2\pi }{26}}={\frac {1+{\sqrt {13}}}{12}}+{\frac {1}{6}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}+i{\frac {3{\sqrt {39}}}{2}}}}+{\frac {1}{6}}{\sqrt[{3}]{{\frac {-(26+5{\sqrt {13}})}{2}}-i{\frac {3{\sqrt {39}}}{2}}}}}$

• 十三角形